An important topic with application in problems on trains, boats and streams, circular motion etc. The entire topic revolves around one formula which is based on the relation between time, speed and distance.
The first section in this topic is an introduction where we look at the formula that explains the relation between time, speed and distance. We will also look at the various proportionals that can be derived from the formula. The exercises contain solved question and answers on time, speed and distance. When solving exercises, we recommend you to try solving the problems orally and with out using equations.
Introduction to Time, Speed and Distance
Relation Between Time, Speed and Distance
Speed = $\frac{\text{Distance Travelled}}{\text{Time Taken}}$
Unit Conversion
- To convert speed in km/hr to m/sec, multiply it with $\frac{5}{18}$
- To convert speed in m/sec to km/hr, multiply it with $\frac{18}{5}$
Identifying Proportionals
If A and B are two objects travelling at a speed
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- Distance is constant:
Speed is inversely proportional to Time. i.e if speed increases time taken to cover the same distance decreases.
Hence, if the ratio of speeds of A and B is s1 : s2, then the time taken by A and B to cover the same distance is in the ratio s2 : s1.
- Time is constant:
Speed is directly proportional to the distance travelled.
Hence, if two objects A and B travel a distance of D1 and D2 with speeds S1 and S2 respectively, then
$\frac{S1}{S2} = \frac{D1}{D2}$
- Speed is constant:
Distance travelled is directly proportional to the time taken.
Hence, if the time taken by A and B to cover a distance of D1 and D2 is T1 and T2 respectively, then
$\frac{T1}{T2} = \frac{D1}{D2}$
Average Speed
Average speed given by Total Distance travelled by Total time taken.
- If a man covers certain distance at u km/hr and an equal distance at v km/hr, then the average speed during the whole journey is given by
$\frac{2uv}{u+v}$ km/hr
- If a person travels for half the time at a speed of u km/hr and remaining half the time at v km/hr, then the average speed is given by
$\frac{u + v}{2}$ km/hr
Problems
Time Speed and Distance – Exercise – Blog
12 questions
-
If a person travels at a speed of 20 km/hr in 10 min find the distance travelled by him.
Show the answer
Answer: 4 kms
Distance travelled = Speed x Time
=$\frac{\text{24 x 10}}{60} ${10 minutes expressed in terms of hours}
= 4 kms
-
If A covers 10 kms in 30 min, find the time taken by A to cover 40 kms at the same speed.
- 2 hours
- 1 hour
- 3 hours
- 5 hours
Show the answer
Answer: 2 hours
Since speed is constant, time taken varies directly as distance travelled.
Let D1 = 10 kms, T1 = 30 min, D2 = 40 kms, T2 = t
$\frac{D1}{T1} =\frac{D2}{T2}$
$\frac{10}{30} =\frac{40}{t}$
t = 120 min or 2 hours
Therefore, time taken by A to cover 40 kms = 2 hours
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Anil runs a certain distance at 10 km/hr. Advith runs the same distance at 12 km/hr. If Anil and advith start running from A to B at the same time and Advith reaches B 15 minutes before Anil. Find the distance between A and B.
Show the answer
Answer: 15 kms
Let the distance between A and B be x.
Anil’s speed, a = 10 km/hr
Advith’s speed, b = 12 km/hr
Time taken =$\frac{\text{Distance travelled}}{Speed}$
Given, Time taken by Advit to reach B – Time taken by Anil to reach B = 15/60.
$\frac{x}{a} -\frac{x}{b} =\frac{15}{60}$
$\frac{x}{10} -\frac{x}{12} =\frac{1}{4}$
x = 15 kms
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Santy travels 100 kms at a speed S. What is the distance covered by Santy if his speed is increased by 50% in the same amount of time.
- 75 kms
- 125 kms
- 150 kms
- 100 kms
Show the answer
Answer: 150 kms
Let D1 = 100 kms, S1 = S, D2 = d and S2 = S (1 + 50/100) = 1.5S
Given time is constant, Distance varies directly as Speed.
Since speed increases by 50%, distance covered will also increase by 50%.
Hence the new distance travelled by Santy at 1.5S is 100 x (1 + $\frac{50}{100}$) = 150 kms
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Shiva travelling at 50% more than his usual speed takes 12 minutes less to cover a distance of 6 km. Find Shiva’s normal speed.
- 60 minutes
- 36 minutes
- 45 minutes
- 54 minutes
Show the answer
Answer: 36 minutes
Let Shiva’s original speed by s. Then, Shiva’s new speed = s (1+$\frac{50}{100}) = 1.5s$
Time travelled at speed s = t minutes
Time travelled at speed 1.5s = t – 12
Since D=500m is constant,
s x t = 1.5s (t -12)
t = 36 minutes
Now,
s x 36 = 6
s =$\frac{6}{36}$ x 60 (From km/minute to km/hr)
= 10 km/hr
-
If Ajith drives at 30 km/hr he will reach 5 minutes late to the office and if he drives at 40 km/hr he will reach 5 minutes early. Find the distance to office.
Show the answer
Answer: 20 kms
Let the distance to office be D. Then,
$\frac{D}{30} -\frac{D}{40} =\frac{10}{60}$ {All the values expressed in terms of kms and hours}
D =$\frac{120}{6}$
= 20 kms
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Chaitanya travelling at a certain speed ‘S’ will be late to his office by 30 mins and if drives at double the speed he will reach 30 minutes hour early. If the office is at distance of 24 kms. Find S in km/hr.
- 8 km/hr
- 12 km/hr
- 6 km/hr
- 10 km/hr
Show the answer
Answer: 12 km/hr
Given,
$\frac{24}{S} -\frac{24}{2S} =\frac{60}{60} ${All the values expressed in terms of kms and hours}
48 – 24 = 2S
S = 12 km/hr
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Ram travels 100 kms in 2 hours by travelling certain distance by bus at 40 km/hr and the remaining distance by car at 60 km/hr. Find the distance travelled by bus.
Show the answer
Answer: 40 kms
Let the distance travelled by bus be x kms. Then,
$\frac{x}{40} +\frac{\text{100-x}}{60} = 2$
3x + 200 – 2x = 240
x = 40 kms
Therefore, distance travelled by bus = 40 kms
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A and B start from opposite ends X and Y towards Y and X respectively at the same time. They take 36 and 25 minutes respectively after they meet at a point Z to reach their destinations. Find the ratio of speeds of A and B.
- $\frac{4}{5}$
- $\frac{5}{4}$
- $\frac{3}{4}$
- $\frac{4}{3}$
Show the answer
Answer: $\frac{5}{4}$
When two persons A and B start at the same time from opposite ends and if the time taken by them to reach respective ends after they meet each other is ta and tb respectively. Then, the ratio of speeds of A and B is $\sqrt{\frac{tb}{ta}}$
Therefore ratio of speeds of A and B $\sqrt{\frac{25}{16}} = \frac{5}{4}$
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A person travels from A to B at a speed of 30 km/hr and returns back to A at a speed of 40 km/hr. Find his average speed.
- 43.28 kms/hr
- 34.28 kms/hr
- 25.56 kms/hr
- 52.56 kms/hr
Show the answer
Answer: 34.28 kms/hr
Here, u = 30 and v = 40.
Average speed = $\frac{2uv}{u+v}$
= $\frac{\text{2 x 30 x 40}}{30 + 40}$
= 34.28 kms/hr
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A man travels one third the distance at 20 km/hr, the next one third the distance at 30 km/hr and the remaining distance at 40 km/hr. Find his average speed.
- $\frac{1800}{37}$ kms/hr
- $\frac{900}{19}$ kms/hr
- 48 kms/hr
- 84 kms/hr
Show the answer
Answer: $\frac{1800}{37}$ kms/hr
Let the distance be equal to 3$x$ kms.
Hence,
The time taken for the first one-third distance = $\frac{x}{20}$
The time for the second one-third distance = $\frac{x}{30}$
The time for the third one-third distance = $\frac{x}{40}$.Total time is the sum of these three.
i.e $\frac{x}{20}$ + $\frac{x}{30}$ + $\frac{x}{40}$ = $\frac{15x+12x+10x}{600}$ = $\frac{37x}{600}$
Average speed equals total distance divided by total timeHence, Average speed = $\frac{3x}{37x/600}$ = $\frac{1800}{37}$ kms/hr
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A person travels the first one hour by Bus at 30 km/hr, the next one hour by Car at 50 km/hr and the last one hour by walk at 10km/hr. Find his average speed.
- 90 kms/hr
- 60 kms/hr
- 45 kms/hr
- 30 kms/hr
Show the answer
Answer: 30 kms/hr
Distance travelled by bus = 1 hr x 30 kms/hr = 30 kms
Distance travelled by car = 1 hr x 50 kms/hr = 50 kms
Distance travelled by walk = 1 hr x 10 kms/hr = 10 kms
Total distance travelled = 30+50+10 = 90 kms
Average speed = total distance by total time
Hence, Average speed = $\frac{90}{3}$ = 30 kms/hr
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